$A$ straight horizontal conducting rod of length $L$ and mass $M$ is suspended by two vertical wires at its ends. If $I$ is the current passing through the rod,then in order that the tension in the wires is zero,the magnetic field set up normal to the conductor is (Neglect the mass of the wire,$g$ = acceleration due to gravity).
- A$\frac{IL}{Mg}$
- B$\frac{Mg}{IL^2}$
- C$\frac{Mg}{I^2 L}$
- D$\frac{Mg}{IL}$
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